Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n}{(\ln n)^n}. $$
Explanation of Method
The root test gives $\sqrt[n]{b_n}=\dfrac{\sqrt[n]n}{\ln n}$. Since $\sqrt[n]n\to1$ while $\ln n\to\infty$, the ratio collapses to $0$, well below the convergence threshold of $1$ — the denominator $(\ln n)^n$ grows far too fast for the slowly growing numerator $n$ to compete.
- State Cauchy's root test.
- Simplify $\sqrt[n]{b_n}$ using $\sqrt[n]n\to1$.
- Conclude $\rho=0<1$, so the series converges.
Step 1 · State the root test
For $b_n\ge0$, let $\rho=\limsup_{n\to\infty}\sqrt[n]{b_n}$: if $\rho<1$ the series converges, if $\rho>1$ it diverges, if $\rho=1$ the test is inconclusive.
Step 2 · Compute $\sqrt[n]{b_n}$
For $n\ge2$, with $b_n=\dfrac n{(\ln n)^n}$:
$$ \sqrt[n]{b_n} = \frac{\sqrt[n]n}{\ln n}. $$
Since $n^{1/n}=e^{(\ln n)/n}$ and $\frac{\ln n}n\to0$, we get $n^{1/n}=1+\frac{\ln n}n+O\!\left(\frac{(\ln n)^2}{n^2}\right)\to1$.
Step 3 · Evaluate the limit and apply the test
$$ \rho = \lim_{n\to\infty}\frac{n^{1/n}}{\ln n} = \lim_{n\to\infty}\left(\frac1{\ln n}+\frac1n\right) = 0. $$
Since $\rho=0<1$, by Cauchy's root test the series converges.
Summary Table
| Quantity | Value |
|---|---|
| $\lim_{n\to\infty}\sqrt[n]n$ | $1$ |
| $\lim_{n\to\infty}\ln n$ | $\infty$ |
| $\rho=\lim\sqrt[n]{b_n}$ | $0$ |
| Conclusion | Converges |
Additional Notes
- The same standard limit $\sqrt[n]n\to1$ used in Problem 70 reappears here — it is the single most useful auxiliary fact for root-test problems involving $n$ in the numerator.
- Unlike Problem 70, where the denominator was exponential ($a^n$) and gave a finite ratio $1/a$, here the denominator $(\ln n)^n$ grows so fast that even the extremely slow-growing $\ln n$, once raised to the power $n$, forces $\rho$ all the way down to $0$.
- This illustrates a general hierarchy: any denominator of the form $c_n^n$ with $c_n\to\infty$ (however slowly) forces $\rho=0$, since $\sqrt[n]{b_n}\sim\frac{\text{poly}(n)^{1/n}}{c_n}\to\frac1{c_n}\to0$.
Result
The series converges: